Answer to Question #105106 in Abstract Algebra for Sourav Mondal

Question

Consider the ideal I=<x³-1, 2x⁴+2x³+7x²+5x+ 5>in Q[x]
Find
p(x) belongs to Q[x]such that I=<p(x)> .
Is Q[x]/I is field ? Give reasons for your answer

Answer 1

$p(x)=(x^3-1,2x^4+2x^3+7x^2+5x+5)$ is the greatest common divisor of 

$x^3-1$ and $2x^4+2x^3+7x^2+5x+5$ .

Find $p(x)$ according to Euclid's algorithm

$2x^4+2x^3+7x^2+5x+5:x^3-1$

$\begin{matrix} 2x^4+2x^3+7x^2+5x+5 &&&& |x^3-1 \\ 2x^4-2x &&&& 2x+2\\ &&2x^3+7x^2+5x+5\\ &&2x^3-2&\\ &&&7x^2+7x+7\\ &&&x^2+x+1 \end{matrix}$

$x^3-1:x^2+x+1$

$\begin{matrix} x^3-1 &&& |x^2+x+1 \\ x^3+x^2+x &&&x-1 \\ & -x^2-x-1\\ &-x^2-x-1\\ &&0 \end{matrix}$

Then $p(x)=x^2+x+1$ .

$Q[x]/p(x)=Q[x] /<x^2+x+1>$ .

The set $Q[x]/p(x)$  the product of polynomials from $Q[x]$ by $p(x)=x^2+x+1$ .

We are obsessed with polynomials $Q[x]$ .

Then $Q[x]/ p(x)=Q[x]/Q[x]=\{0\}$

It is not a field, because there  $x^2+x+1=0$

Answer: $p(x)=x^2+x+1$ , $Q[x]/I$ is not a field